Find the area of the parallelogram with vertices A(−1,2,3), B(0,4,6), C(1,1,2), and D(2,3,5).
cupoosta [38]
Answer:
5*sqrt3
Step-by-step explanation:
The vector AB= (0-(-1), 4-2,6-3) AB= (1,2,3)
The modul of AB is sqrt(1^2+2^2+3^2)= sqrt14
The vector AC is (1-(-1), 1-2, 2-3)= (2,-1,-1)
The modul of B is sqrt (2^2+(-1)^2+(-1)^2)= sqrt6
AB*AC= modul AB*modul AC*cosA
cosA=( 1*2+2*(-1)+3*(-1))/ sqrt14*sqrt6= -3/sqrt84=
sinB= sqrt (1- (-3/sqrt84)^2)= sqrt75/84= sqrt 25/28= 5/sqrt28
s= modul AB*modul AC*sinA= sqrt14*sqrt6* 5/ sqrt28= 5*sqrt3
Answer:
xcvbxc vx
Step-by-step explanation:
xcv xc c cx xc xcz xzcxcxzc
9514 1404 393
Answer:
10
Step-by-step explanation:
If there is a constant rate of change, its value will be ...
k = y/x
k = 250/25 = 10
The constant rate of change is 10.
y = 10x
-40
Explanation:
(-7+2) would equal -5, then you multiply -5 by 8 and you get -40.
Answer:
16
Step-by-step explanation:
Start by adding 9 to both sides, obtaining: √(x + 9) = 5.
Square both sides, obtaining: x + 9 = 25.
Subtract 9 from both sides: x = 16
Note that √(16+9) - 9 = -4, as required.
